【C/PTA —— 10.函数1（课外实践）】

一.函数题

6-1 符号函数

int sign(int x)
{
    
    
    if (x > 0)
        return 1;
    else if (x == 0)
        return 0;
    else
        return -1;
}

6-2 求排列数

double fact(int n)
{
    
    
    long long ret = 1;
    for (long long i = 1; i <= n; i++)
    {
    
    
        ret =ret * i;
    }
    return ret;
}

6-3 求一个大于10的n位整数w的后n-1位的数，并作为函数值返回。

#include<math.h>
int fun(int w)
{
    
    
	int sum = 0;
	int count = 0;
	int num = w;
	while (num)
	{
    
    
		num /= 10;
		count++;
	}
	sum = w % (int) (pow(10, count - 1));
	return sum;
}

6-4 其右上三角（含主对角线）元素之和。

int fun(int a[3][3])
{
    
    
    int sum1 = 0, sum2 = 0;
    for (int i = 0; i < 3; i++)
    {
    
    
        for (int j = i; j < 3; j++)
        {
    
    
            sum1 += a[i][j];
        }
    }
    return sum1;
} 

6-5 字符串比较

int fun(char a[], char b[])
{
    
    
	if (strcmp(a, b) < 0)
		return -1;
	else if (strcmp(a, b) > 0)
		return 1;
	else
		return strcmp(a, b);
}

6-6 使用函数求素数和

int prime(int p)
{
    
    
    int count = 0;
    if(p==1)
        return 0;
    for (int i = 2; i <= p; i++)
    {
    
    
        if (p % i == 0)
            count++;
    }
    if (count == 1)
        return 1;
    else
        return 0;
}
int PrimeSum(int m, int n)
{
    
    
    int sum = 0;
    for (int i = (m<=n?m:n); i <= (n>=m?n:m); i++)
    {
    
    
        if (prime(i) == 1)
            sum += i;
    }
    return sum;
}

6-7 使用函数输出水仙花数

#include<math.h>
int narcissistic(int number)
{
    
    
    int count = 0;
    int sum1 = 0;
    int num = number;
    while (number)
    {
    
    
        number /= 10;
        count++;
    }
    int sum = 0;
    number = num;
    while (number)
    {
    
    
        sum1 = number % 10;
        sum += (int)(pow(sum1, count));
        number /= 10;
        sum1 = 0;
    }
    if (num == sum)
        return 1;
    else
        return 0;
}
void PrintN(int m, int n)
{
    
    
    for (int i = (m <= n ? m : n)+1; i < (n >= m ? n : m); i++)
    {
    
    
        if (narcissistic(i) == 1)
        {
    
    
            printf("%d\n", i);
        }
    }
} 

二.编程题

7-1 寻找自守数

#include<stdio.h>
int ZISHOUnum(int p)
{
    
    
	int num = 0, count = 0;
	num = p * p;
	while (p != 0)
	{
    
    
		if (p % 10 != num % 10)
			return 0;
		p /= 10;
		num /= 10;
	}
	return 1;
}

int main()
{
    
    
	int n = 0, m = 0;
	int count = 0;
	scanf("%d %d", &n, &m);
	for (int i = n; i <= m; i++)
	{
    
    
		if (ZISHOUnum(i))
		{
    
    
			count++;
			printf("%d\n", i);
		}
			
	}
	if (count == 0)
	{
    
    
		printf("None\n");
	}
	return 0;
}